1.) Water flows from a 2 cm diameter pipe, at a speed of 0.35m/sec. How long will it take to fill a 10 liter container?
2.) A horizontal segment of pipe tapers from a cross-sectional area of 50 cm^2 to 0.5 cm^2. The pressure at the larger end of the pipe is 1.2x 10^5 Pa and the speed is 0.04m/sec. What is the pressure at the narrow end of the segment?
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1.) To find the time it will take to fill a 10 liter container, we need to know the volume of water that flows through the pipe per second.
First, we need to calculate the cross-sectional area of the pipe. The formula for the area of a circle is A = πr^2, where r is the radius. Since the diameter of the pipe is given as 2 cm, the radius is 1 cm or 0.01 m.
So, the cross-sectional area is A = π(0.01 m)^2 = 0.000314 m^2.
Next, we need to find the volume of water that flows through the pipe per second. The formula for the volume flow rate is Q = A × v, where Q is the volume flow rate, A is the cross-sectional area, and v is the speed of the water flow.
Substituting the values we have, Q = 0.000314 m^2 × 0.35 m/sec = 0.0001099 m^3/sec.
Now, we need to convert the volume flow rate to liters per second, as the capacity of the container is given in liters. Since 1 liter is equal to 0.001 m^3, we can convert the volume flow rate as follows:
Q = 0.0001099 m^3/sec × (1/0.001) = 0.1099 liters/sec.
Finally, we can calculate the time it will take to fill a 10 liter container by dividing the capacity of the container by the volume flow rate:
Time = 10 liters / 0.1099 liters/sec ≈ 91.01 seconds. Therefore, it will take approximately 91.01 seconds to fill the 10 liter container.
2.) To find the pressure at the narrow end of the pipe segment, we can use the principle of continuity and Bernoulli's equation.
The principle of continuity states that the mass flow rate of an incompressible fluid is constant along a streamline. This means that the product of the cross-sectional area and the speed of the fluid remains constant.
First, we can calculate the speed of the fluid at the narrow end of the segment using the principle of continuity. The formula is A1v1 = A2v2, where A1 and A2 are the cross-sectional areas at the larger and narrower ends respectively, and v1 and v2 are the speeds at the larger and narrower ends respectively.
Given that A1 = 50 cm^2 = 0.005 m^2, A2 = 0.5 cm^2 = 0.00005 m^2, and v1 = 0.04 m/sec, we can rearrange the equation to find v2:
v2 = (A1v1) / A2 = (0.005 m^2 × 0.04 m/sec) / 0.00005 m^2 ≈ 4 m/sec.
Now, we can use Bernoulli's equation to find the pressure at the narrow end of the segment. The equation is P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2, where P1 and P2 are the pressures at the larger and narrower ends respectively, ρ is the density of the fluid (assumed to be constant), v1 and v2 are the speeds at the larger and narrower ends respectively, g is the acceleration due to gravity, and h1 and h2 are the heights at the larger and narrower ends respectively (which we assume to be the same).
Given that P1 = 1.2x10^5 Pa, v1 = 0.04 m/sec, and v2 = 4 m/sec, we can rearrange the equation to find P2:
P2 = P1 + 0.5ρv1^2 - 0.5ρv2^2 ≈ 1.2x10^5 Pa + 0.5ρ(0.04 m/sec)^2 - 0.5ρ(4 m/sec)^2.
Without knowing the value of the density of the fluid, we cannot determine the exact pressure at the narrow end. However, by following the steps above, we have derived the formula required to calculate it once the density is known.